When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 xx 10^(5)J mol^(-1). What is the minimum energy needed to remove an electron from sodium ? What is the maximum wavelength that will cause a photoelectron to be emitted?

When electromagnetic radiation of wavelength 300 nm falls on the surfa

1.	When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 xx 10^(5)J mol^(-1). What is the minimum energy needed to remove an electron from sodium ? What is the maximum wavelength that will cause a photoelectron to be emitted?
Answer» Solution :Energy of a photon of radiation of wavelength 300 nm will be `E = hv = h (c)/(lamda) = ((6.626 xx 10^(-34) Js) (3.0 xx 10^(8) ms^(-1)))/((300 xx 10^(-9) m)) = 6.626 xx 10^(-19) J` `:.` Energy of 1 mole of PHOTONS `= (6.626 xx 10^(-19)J) xx (6.022 xx 10^(23) mol^(-1)) = 3.99 xx 10^(5) J mol^(-1)` As `E = E_(0) +` K.E. of photoelectrons emitted. `:.` Minimum energy `(E_(0))` required to remove 1 mole of electrons from sodium `= E -` K.E. `= (3.99 - 1.68) 10^(5) J mol^(-1) = 2.31 xx 10^(5) J mol^(-1)` `:.` Minimum energy required to remove one electron `= (2.31 xx 10^(5) J mol^(-1))/(6.022 xx 10^(23) mol^(-1)) = 3.84 xx 10^(-19) J` The wavelength corresponding to this energy can be calculated using the formula, `E = hv = h (c)/(lamda)` `:. lamda = (hc)/(E) = ((6.626 xx 10^(-34) Js) (3.0 xx 0^(8) ms^(-1)))/(3.84 xx 10^(-19)J) = 5.17 xx 10^(-7) m = 517 xx 10^(-9) m = 517 nm` which CORRESPONDS to the green light. As `lamda prop (1)/(E)`, hence when E is minimum, `lamda` is maximum